If the equation of parabola is $y^2 = 8x$, then locus of $P$ is:
Step-by-Step Solution
Key Concept: For a parabola y² = 8x (with parameter a = 2), normals drawn from external point P(h,k) generate a cubic equation in slope m. Using Vieta's formulas on this cubic, the condition that two normals are perpendicular (m₁m₂ = -1) combined with the third normal's slope constraint yields the locus equation.
For normals to $y^2 = 8x$ passing through point P, the cubic equation $2m^3 + (4-h)m = 0$ arises. By Vieta's formulas with roots $m_1, m_2, m_3$: $m_1m_2 = -1$ and $m_3 = \frac{k}{2}$. Substituting back into the root condition and solving yields $k^2 = 2(h-6)$, giving the locus $y^2 = 2(x-6)$.
Correct Answer: 2